answer：Okay, so I have this problem here: If a is less than b, which of the following expressions must hold true? The options are A, B, C, and D. Let me try to figure this out step by step.First, let's understand what each option is saying.Option A: -a < -b. Hmm, so if a is less than b, does that mean that negative a is less than negative b? Wait, I remember that when you multiply or divide both sides of an inequality by a negative number, the inequality sign flips. So if a < b, then multiplying both sides by -1 would give -a > -b. That means -a is actually greater than -b, not less. So option A is incorrect. Got it.Option B: a - 1 < b - 1. Okay, so if we subtract 1 from both a and b, does the inequality still hold? Let me think. If a is less than b, then subtracting the same number from both sides shouldn't change the inequality. For example, if a is 2 and b is 3, then 2 - 1 is 1 and 3 - 1 is 2, so 1 < 2, which is true. Another example: if a is -1 and b is 0, then -1 -1 is -2 and 0 -1 is -1, so -2 < -1, which is also true. It seems like subtracting the same number from both sides preserves the inequality. So option B is correct.Option C: |a/3| > |b/3|. Alright, so the absolute value of a third of a is greater than the absolute value of a third of b. Let me consider some numbers. Suppose a is 1 and b is 2. Then |1/3| is 1/3 and |2/3| is 2/3. So 1/3 is not greater than 2/3; it's actually less. So in this case, option C is false. What if a and b are negative? Let's say a is -4 and b is -3. Then |-4/3| is 4/3 and |-3/3| is 1. So 4/3 is greater than 1, which would make option C true in this case. Hmm, so sometimes it's true and sometimes it's not. Therefore, option C doesn't have to be true every time. So it's not necessarily correct.Option D: ac < bc. Okay, so if we multiply both sides of the inequality by c, does it hold? Wait, I remember that if you multiply both sides by a positive number, the inequality remains the same, but if you multiply by a negative number, it flips. But the problem is, we don't know what c is. If c is positive, then ac < bc would hold because a < b. But if c is negative, then ac would be greater than bc because multiplying by a negative flips the inequality. For example, let's say a is 1, b is 2, and c is 3. Then 1*3 = 3 and 2*3 = 6, so 3 < 6, which is true. But if c is -1, then 1*(-1) = -1 and 2*(-1) = -2, so -1 is greater than -2, which means ac > bc, so the inequality doesn't hold. Therefore, without knowing the sign of c, we can't be sure if ac < bc is true or not. So option D is not necessarily correct.So, going back through the options:A: Incorrect because flipping the inequality when multiplying by a negative changes the direction.B: Correct because subtracting the same number from both sides preserves the inequality.C: Sometimes correct, sometimes not, depending on the values of a and b, so it's not necessarily true.D: Depends on the sign of c, which we don't know, so it's not necessarily true.Therefore, the only expression that must hold true is option B.

answer：First, I'll define the amounts of money that A, B, and C have as A, B, and C respectively.From the problem, I have three pieces of information:1. The total amount of money that A, B, and C have together is Rs. 450: A + B + C = 4502. A and C together have Rs. 200: A + C = 2003. B and C together have Rs. 350: B + C = 350My goal is to find out how much money C has.Starting with the second equation, I can express A in terms of C:A = 200 - CNext, I'll substitute this expression for A into the first equation:(200 - C) + B + C = 450Simplifying this equation:200 + B = 450Solving for B:B = 450 - 200B = 250Now that I know B is 250, I'll substitute this value into the third equation to find C:250 + C = 350Solving for C:C = 350 - 250C = 100Therefore, C has Rs. 100.

answer：Okay, so I have this equation to solve: x squared minus 4x minus 5 equals zero. Hmm, quadratic equations. I remember they have the standard form ax squared plus bx plus c equals zero. In this case, a is 1, b is -4, and c is -5. Alright, I think there are a couple of ways to solve quadratic equations. One is factoring, another is using the quadratic formula, and maybe completing the square. I'm not sure which one to use here. Let me try factoring first because it seems simpler if it works.Factoring quadratics usually involves finding two numbers that multiply to give the constant term (which is -5 here) and add up to the coefficient of the middle term (which is -4). So, I need two numbers that multiply to -5 and add up to -4. Let me think: factors of -5 could be 1 and -5, or -1 and 5. Let's see, 1 and -5: 1 plus (-5) is -4. Oh, that works! So, that means I can factor the quadratic as (x + 1)(x - 5) equals zero. Now, according to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero. So, I can set each factor equal to zero and solve for x.First factor: x + 1 equals zero. Subtracting 1 from both sides gives x equals -1.Second factor: x - 5 equals zero. Adding 5 to both sides gives x equals 5.So, the solutions are x equals -1 and x equals 5. Wait, let me double-check to make sure I didn't make any mistakes. I'll plug the values back into the original equation to verify.For x equals -1: (-1)^2 minus 4*(-1) minus 5 equals 1 plus 4 minus 5, which is 0. That works.For x equals 5: 5^2 minus 4*5 minus 5 equals 25 minus 20 minus 5, which is also 0. That works too.Okay, so factoring worked out fine here. I guess I didn't need to use the quadratic formula after all. But just in case, let me try it with the quadratic formula to confirm.The quadratic formula is x equals negative b plus or minus the square root of b squared minus 4ac all over 2a. Plugging in the values: b is -4, so negative b is 4. b squared is (-4)^2, which is 16. 4ac is 4*1*(-5), which is -20. So, the discriminant is 16 minus (-20), which is 16 plus 20, so 36. The square root of 36 is 6.So, x equals (4 plus or minus 6) over 2. That gives two solutions: (4 + 6)/2 equals 10/2 equals 5, and (4 - 6)/2 equals (-2)/2 equals -1. Same results as before. So, both methods give me x equals -1 and x equals 5. That makes me confident that these are the correct solutions.I think I'm done here. The solutions are x equals -1 and x equals 5.

answer：First, determine the number of school days Alice walks to school and back home. She walks from Monday to Friday, which is 5 days.Next, calculate the total distance she walks each day. She walks 10 miles to school in the morning and 12 miles back home in the afternoon, totaling 22 miles per day.Finally, multiply the daily distance by the number of school days to find the total distance walked in the week: 22 miles/day × 5 days = 110 miles.

answer：Okay, so I have this problem about an ellipse, and I need to find the distance from a point P on the ellipse to one of the foci, given the distance to the other focus. Let me try to break this down step by step.First, the equation of the ellipse is given as ( frac{x^2}{9} + frac{y^2}{6} = 1 ). I remember that the standard form of an ellipse is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where 'a' is the semi-major axis and 'b' is the semi-minor axis. So in this case, ( a^2 = 9 ) and ( b^2 = 6 ). That means ( a = 3 ) and ( b = sqrt{6} ).Now, the foci of an ellipse are located at a distance 'c' from the center along the major axis. I recall that ( c^2 = a^2 - b^2 ). Let me calculate that. So, ( c^2 = 9 - 6 = 3 ), which means ( c = sqrt{3} ). Therefore, the foci ( F_1 ) and ( F_2 ) are located at ( (-sqrt{3}, 0) ) and ( (sqrt{3}, 0) ) respectively.The problem states that point P lies on the ellipse and the distance from P to ( F_1 ) is 2, i.e., ( |PF_1| = 2 ). I need to find ( |PF_2| ).I remember a key property of ellipses: the sum of the distances from any point on the ellipse to the two foci is constant and equal to the major axis length, which is ( 2a ). In this case, since ( a = 3 ), the sum should be ( 2 times 3 = 6 ).So, if ( |PF_1| + |PF_2| = 6 ), and ( |PF_1| = 2 ), then ( |PF_2| = 6 - 2 = 4 ).Wait, that seems straightforward. But let me double-check if I applied the property correctly. The major axis is along the x-axis because the larger denominator is under ( x^2 ), so the foci are indeed on the x-axis. The distance from P to each focus adds up to the major axis length, which is 6. Subtracting the given distance of 2 gives me 4 for the other distance. That makes sense.Is there any chance I made a mistake in calculating 'c'? Let me verify. ( c^2 = a^2 - b^2 = 9 - 6 = 3 ), so ( c = sqrt{3} ). Yes, that's correct. So the foci are correctly placed at ( pm sqrt{3}, 0 ).Another thing to consider: could there be a case where the point P is such that it's closer to one focus and farther from the other? But in an ellipse, regardless of where P is, the sum remains constant. So even if P is near ( F_1 ), the sum ( |PF_1| + |PF_2| ) is still 6.Therefore, I think my reasoning is solid. The distance from P to ( F_2 ) should be 4.

answer：Okay, so I have this inequality system to solve, and I need to find the range of values for 'a' such that the solution set is x < 1. The system is:1. x - 1 < 02. x < a + 3Alright, let me start by understanding each inequality separately.First, the first inequality is x - 1 < 0. If I solve this, I can add 1 to both sides, which gives me x < 1. So, from the first inequality alone, I know that x has to be less than 1.Now, the second inequality is x < a + 3. This one is a bit trickier because it involves 'a', which is a variable we need to find the range for. So, I need to figure out what values of 'a' will make sure that when combined with the first inequality, the solution set remains x < 1.Let me think about how these two inequalities interact. The first one tells me x has to be less than 1, and the second one tells me x has to be less than a + 3. So, the solution set of the system will be the overlap of these two conditions.If I imagine a number line, the first inequality restricts x to the left of 1, and the second inequality restricts x to the left of a + 3. So, the more restrictive condition will determine the solution set.Given that the solution set is x < 1, this means that a + 3 must be greater than or equal to 1. If a + 3 were less than 1, then the solution set would be x < a + 3, which would be more restrictive than x < 1, and that's not what we want. We want the solution set to be exactly x < 1, so a + 3 needs to be at least 1.Let me write that down:a + 3 ≥ 1Now, solving for 'a', I subtract 3 from both sides:a ≥ 1 - 3a ≥ -2So, 'a' must be greater than or equal to -2.Wait, let me double-check that. If a is -2, then a + 3 is 1, which means the second inequality becomes x < 1, which is the same as the first inequality. So, the solution set is x < 1, which matches what we need.If a is greater than -2, say a = 0, then a + 3 is 3, so the second inequality is x < 3. But since the first inequality is x < 1, the solution set is still x < 1 because 1 is less than 3. So, that works too.But if a is less than -2, say a = -3, then a + 3 is 0, so the second inequality is x < 0. Now, the solution set would be x < 0 because that's more restrictive than x < 1. But we need the solution set to be x < 1, so a can't be less than -2.Therefore, the range of values for 'a' is all real numbers greater than or equal to -2.I think that makes sense. Let me summarize:1. Solve the first inequality: x < 1.2. The second inequality is x < a + 3.3. For the solution set to be x < 1, a + 3 must be ≥ 1.4. Solving a + 3 ≥ 1 gives a ≥ -2.So, the range of values for 'a' is a ≥ -2.